CF1106E Lunar New Year and Red Envelopes

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传送门

容易发现对于第 ii 秒，Bob 的选择是固定的，所以可以用线段树 O(nlog⁡n)O(n\log n) 预处理出来 Bob 每一秒选择的 wiw_i 和 did_i。

然后考虑倒着 dp，我们设 fi,jf_{i,j} 表示第 ii 到 nn 秒干扰 jj 次收集到的最少钱数，转移方程就是：

fi,j=min⁡(fdi+1,j+wi,fi+1,j−1)f_{i,j}=\min(f_{d_i+1,j}+w_i,f_{i+1,j-1})

时间复杂度 O(nm)O(nm)。

Code

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#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e5 + 5;
int n, m, k;
struct segment_tree
{
    int tagd[maxn << 2], tagw[maxn << 2];
    void push_down(int p)
    {
        if (tagw[p] > tagw[p * 2])
            tagw[p * 2] = tagw[p], tagd[p * 2] = tagd[p];
        else if (tagw[p] == tagw[p * 2] && tagd[p] > tagd[p * 2])
            tagd[p * 2] = tagd[p], tagw[p * 2] = tagw[p];
        if (tagw[p] > tagw[p * 2 + 1])
            tagw[p * 2 + 1] = tagw[p], tagd[p * 2 + 1] = tagd[p];
        else if (tagw[p] == tagw[p * 2 + 1] && tagd[p] > tagd[p * 2 + 1])
            tagd[p * 2 + 1] = tagd[p], tagw[p * 2 + 1] = tagw[p];
    }
    void modify(int p, int l, int r, int nl, int nr, int d, int w)
    {
        if (l >= nl && r <= nr)
        {
            if (tagw[p] < w)
                tagw[p] = w, tagd[p] = d;
            else if (tagw[p] == w && tagd[p] < d)
                tagw[p] = w, tagd[p] = d;
            return;
        }
        int mid = (l + r) / 2;
        push_down(p);
        if (mid >= nl)
            modify(p * 2, l, mid, nl, nr, d, w);
        if (mid < nr)
            modify(p * 2 + 1, mid + 1, r, nl, nr, d, w);
    }
    pii query(int p, int l, int r, int x)
    {
        if (l == r)
            return {tagw[p], tagd[p]};
        int mid = (l + r) / 2;
        push_down(p);
        if (mid >= x)
            return query(p * 2, l, mid, x);
        else
            return query(p * 2 + 1, mid + 1, r, x);
    }
} T;
int d[maxn], w[maxn];
ll f[maxn][205];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> n >> m >> k;
    for (int i = 1; i <= k; i++)
    {
        int l, r, s, t;
        cin >> l >> r >> s >> t;
        T.modify(1, 1, n, l, r, s, t);
    }
    for (int i = 1; i <= n; i++)
    {
        pii tmp;
        tmp = T.query(1, 1, n, i);
        d[i] = tmp.second;
        w[i] = tmp.first;
    }
    memset(f, 0x3f, sizeof(f));
    f[n][0] = w[n], f[n][1] = 0, f[n + 1][0] = 0;
    for (int i = n - 1; i >= 1; i--)
    {
        if (!w[i])
        {
            for (int j = 0; j <= m; j++)
                f[i][j] = f[i + 1][j];
            continue;
        }
        for (int j = 0; j <= m; j++)
        {
            f[i][j] = min(f[d[i] + 1][j] + w[i], f[i][j]);
            if (j != 0)
                f[i][j] = min(f[i][j], f[i + 1][j - 1]);
        }
    }
    ll ans = 0x3f3f3f3f3f3f3f3f;
    for (int i = 0; i <= m; i++)
        ans = min(ans, f[1][i]);
    cout << ans << endl;
    return 0;
}